Question

The enthalpy of neutralization of a strong acid by a strong base is $-57.32{\text{kJ mol}}^{-1}$. The enthalpy of formation of water and hydrogen ions are $-285.84{\text{kJ mol}}^{-1}$ and $0{\text{kJ mol}}^{-1}$ respectively. The enthalpy of formation of the hydroxyl ion is:

• A. $228.52{\text{kJmol}}^{-1}$
• B. $-114.26{\text{kJmol}}^{-1}$
• C. $-228.52{\text{kJmol}}^{-1}$
• D. $114.2{\text{kJmol}}^{-1}$

Answer 1

The correct option is C $-228.52{\text{kJmol}}^{-1}$

The enthalpy of neutralization is:
$H^{+}\left(aq\right)+O{H}^{-1}\left(aq\right)\to H_{2}O\left(l\right);\Delta H^{\circ }=-57.32{\text{kJmol}}^{-1}$
We know,
$\Delta H_{reaction}=\Sigma \left(\text{heat of formation of products}\right)-\Sigma \left(\text{heat of formation of reactants}\right)$
$\Rightarrow -57.32=\Delta H_{f_{H_{2}O\left(l\right)}}^{\circ }-[\Delta H_{f_{\left(H^{+}\right)\left(aq\right)}}^{\circ }+\Delta H_{f_{\left(O{H}^{-}\right)\left(aq\right)}}^{\circ }]$
$\Rightarrow -57.32=-285.84-(0+x)$
$\Rightarrow x=-283.84+57.32=-228.52{\text{kJmol}}^{-1}$